Question

How much energy is required to change 300.0 g of water from 20.0°C to 45.0°C? (C-water = 4.18 J/g • °C / Q = mc ΔT)

Answer 1

To calculate the energy required to change the temperature of water, you can use the formula:

\[
Q = mc \Delta T
\]

Where:
- \( Q \) is the heat energy (in joules),
- \( m \) is the mass of the substance (in grams),
- \( c \) is the specific heat capacity (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).

Given:
- \( m = 300.0 \, \text{g} \)
- \( c = 4.18 \, \text{J/g°C} \)
- Initial temperature = \( 20.0 \, \text{°C} \)
- Final temperature = \( 45.0 \, \text{°C} \)

First, calculate the change in temperature (\( \Delta T \)):

\[
\Delta T = \text{Final Temperature} - \text{Initial Temperature} = 45.0 \, \text{°C} - 20.0 \, \text{°C} = 25.0 \, \text{°C}
\]

Now, plug the values into the equation:

\[
Q = (300.0 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (25.0 \, \text{°C})
\]

Calculating it step-by-step:

\[
Q = 300.0 \times 4.18 \times 25.0
\]

\[
Q = 300.0 \times 104.5
\]

\[
Q = 31350 \, \text{J}
\]

Thus, the energy required to change 300.0 g of water from 20.0°C to 45.0°C is **31,350 J**.